User alessandro sisto - MathOverflow

Answer by Alessandro Sisto for Recognising group actions on trees from the boundary

2012-05-18T13:36:05Z

You may wish to take a look at "Quasi-actions on trees. I." by Mosher, Sageev and Whyte. They consider cobounded quasi-actions by quasi-isometries with bounded constants on trees with bounded valence and show that they are conjugate to actual isometric actions on trees. As noted in Theorem 5 of the same paper, such actions correspond to uniformly quasiconformal actions on the boundary which are cocompact on the space of triples.

Centralizers of non-iwip elements of $Out(F_n)$

2011-12-07T17:42:43Z

Does there exist an infinite order element $\phi\in Out(F_n)$ , for some or all $n\geq 3$ , which is not iwip but has finite index in its centralizer? How about an element such that all its non-zero powers have this property?

Motivation: iwip elements are Morse (i.e., roughly speaking, all quasi-geodesics connecting points on a given orbit stay close to it), and a standard way of proving that an element is not Morse is showing that it has infinite index in its centralizer, or one of its powers do. An element in the mapping class group of a closed surface is Morse if and only if it is pseudo-Anosov.

Asymptotic dimension of graph manifold groups

2011-11-30T14:00:19Z

Does every non-geometric graph manifold have fundamental group of asymptotic dimension 3?

This is affirmed in http://arxiv.org/abs/0909.1098 for closed graph manifolds, but I am interested in non-closed graph manifolds as well. Notice that the asymptotic dimension of such groups is always at least 2 (obvious) and at most 3 (by a result of Bell and Dranishnikov).

Answer by Alessandro Sisto for Asymptotic dimension of graph manifold groups

2012-03-06T14:25:12Z

I should have seen much earlier that it's 2. In fact, every non-closed graph manifold has a finite sheeted cover that fibers over the circle, see here. The fundamental group of such cover is an HNN extension of a free group, so that it has asymptotic dimension at most 2 by the main result of this paper.

Answer by Alessandro Sisto for A question about compact subsets of Hilbert space

2011-12-12T16:10:44Z

Instead of using compact subsets one can just use finite subsets. Now, consider any finite subset of $B$ : it is contained in a finite dimensional subspace $V$ . There is a unit vector perpendicular to $V$ , and such vector has distance at least $1$ from the given finite subset.

Answer by Alessandro Sisto for Local finiteness and coarse bounded geometry

2011-09-09T15:39:52Z

The answer to question 2 is negative, but if you require quasi-homogeneity (i.e. you have a group of isometries with a $c-$ dense orbit form some $c$ ) then it becomes affirmative. You typically have this.

Also, to construct examples as in question 1 you can consider non-quasi-homogeneous spaces. Hope this helps, I can be more explicit on this point if you need clarifications.

Answer by Alessandro Sisto for Regularity of asymptotic cones

2010-11-15T00:07:22Z

I would just like to add to the answer by Simon Thomas that

-if a group is virtually nilpotent, its asymptotic cones are very regular: they have a Lie group structure and their metric is of Carnot-Caratheodory type (these metrics are described in the wikipedia article "Sub-Riemannian manifold"). Also, the asymptotic cones do not depend on the scaling factor.

-if a group is not virtually nilpotent, its asymptotic cones tend to be VERY large objects. For example, the asymptotic cones of each non-virtually cyclic hyperbolic group are real trees with valency $2^{\aleph_0}$ at each point (those groups have exponential growth, I have to admit that I know very little about asymptotic cones of groups of intermediate growth).

Answer by Alessandro Sisto for Monotone injection of an ordinal into $[0,1]$

2010-10-09T14:44:44Z

No, because you could use it to construct an injective map $\omega_1\to\mathbb{Q}$ , mapping $\alpha<\omega_1$ to some rational number between $\alpha$ and `$\alpha+1$.

Answer by Alessandro Sisto for When completion of locally compact length space is locally compact?

2010-09-30T20:31:42Z

A necessary and sufficient condition (but I do not feel satisfied with that) for the locally compact length space $X$ to have a locally compact completion is that there exists some $r>0$ such that each ball of radius $r$ in $X$ is totally bounded.

In fact, if the condition holds closed balls of radius $r/2$ in $\overline{X}$ are compact. On the other hand, suppose that $\overline{X}$ is locally compact. Then, as it is a complete length space, it is proper (this is called the Hopf-Rinow Theorem in the book by Bridson and Haefliger). This should imply that balls of any radius in $X$ are totally bounded.

The main reason why I am not satisfied with it is that the proof that the condition is sufficient does not use that $X$ is a length space, so this is not really the answer to what you asked. I thought it might be relevant, anyway...

Answer by Alessandro Sisto for Distinct well-orderings of the same set

2010-09-29T19:36:04Z

Here is a quick sketch (probably it can be made much cleaner). Using an inductive argument it should not be difficult to reduce to studying the case that $(X,<_1)$ is isomorphic to a cardinal, say $\kappa$. For convenience, let us identify $(\kappa,<)$ and $(X,<_1)$ . Let us now construct $Y$ using transfinite induction. Let $X'\subseteq X$ be the initial segment of $X$ (with respect to $<_2$ ) which, endowed with the order $<_2$ , is isomorphic to $\kappa$. For $\alpha<k$ define $$y_\alpha=\min\{\beta\in X': \forall \alpha'<\alpha\; \beta> y_{\alpha'}\textrm{\ and\ }\beta>_2 y_{\alpha'}\}.$$ The set above is not empty, and so $y_\alpha$ is well defined, because $|\{\gamma\in X':\exists \alpha'<\alpha\; \gamma\leq y_{\alpha'}\}|<k$ and also $|\{\gamma\in X':\exists \alpha'<\alpha\; \gamma\leq_2 y_{\alpha'}\}|<k$ (this depends on $\{y_{\alpha'}\}_{\alpha'<\alpha}$ not being cofinal in $(k,<)$ and $(X',<_2)$ ), so their complements in $X'$ intersect. The set $Y=\{y_\alpha\}_{\alpha<k}$ is what we were looking for.

Comment by Alessandro Sisto

2012-03-28T22:20:15Z

This appears as a problem here: sci.ccny.cuny.edu/~shpil/gworld/problems/… (Assuming that the class number is the number of conjugacy classes)

Comment by Alessandro Sisto

2011-12-09T00:40:38Z

@Ashot: you're right! Is it easy to see that your elements actually have finite index in their centralizers?

Comment by Alessandro Sisto

2011-12-08T16:13:15Z

I think that elements fixing the first 3 generators and conjugating the fourth one are in the centralizer of all the outer automorphisms you mention, unfortunately.

Comment by Alessandro Sisto

2011-12-07T21:34:57Z

An element of $Out(F_n)$ is irreducible if it does not fix (the conjugacy class of) some free factor $F<F_n$ (i.e. $F_n=F*G$ for some other subgroup $G$ ). It is irreducible with irreducible powers (iwip) if its powers have the same property as well.

Comment by Alessandro Sisto

2011-12-07T21:31:04Z

Behrstock proved an equivalent property in this paper (Theorem 6.5): arxiv.org/abs/math/0502367 Another proof can be found here: arxiv.org/abs/0801.4141 The idea of this last proof is best explained pretending that curve complexes are trees. If this was true then all paths connecting markings projecting on distant simplices will have to cross the pre-images in the marking complex of certain specified simplices. An orbit of a pA does this "as fast as possible", and one can show that a path staying far from such orbit can't "go straight" (acylindricity) and so it is much longer.

Comment by Alessandro Sisto

2011-09-10T14:04:51Z

As I mentioned, it means that there are many isometries, meaning that there exists some $c$ such that for each $x,y$ there exists an isometry mapping $y$ to a point at distance at most $c$ from $x$ . Notice that being homogeneous is the same thing as being quasi-homogeneous with constant $c=0$ .

Comment by Alessandro Sisto

2011-09-10T00:00:37Z

@Bill: I think it has been edited. In any case, I removed the "ps" part, thanks for pointing this out.

Comment by Alessandro Sisto

2011-06-24T18:47:40Z

I personally seriously doubt that there exists such an algebraic condition, as there are many manifolds with a given fundamental group and you can perturb the metric on a Riemannian manifold in several ways. I would be very surprised if the property you require was stable under those perturbations (except in the case of $S^1$ ), see the comment by Sergei Ivanov. Btw, the only examples of manifolds with that property I can think of are tori (with flat metric). And simply connected manifolds.

Comment by Alessandro Sisto

2011-04-18T23:24:20Z

@Joe: try removing "?isAuthorized=no". Anyway it's "Topology in General Relativity" by Robert Geroch

Comment by Alessandro Sisto

2011-04-18T21:54:29Z

I find the related problem of what the topology of (space-like sections of) the universe is and whether it can change very interesting. This clearly relates to cobordism theory, see jmp.aip.org/resource/1/jmapaq/v8/i4/…

Comment by Alessandro Sisto

2011-03-07T22:23:11Z

Thanks, I didn't know that result on triangulations.

Comment by Alessandro Sisto

2011-03-07T13:08:18Z

The connected sum you describe is second countable only if there are countably many closed n-manifolds up to diffeomorphism. Please forgive me for my ignorance: is this the case? Why?

Comment by Alessandro Sisto

2011-01-26T21:40:03Z

Thanks Martin, I thought that the starting point was a "random" embedding.

Comment by Alessandro Sisto

2011-01-26T21:22:54Z

There must be something I am misunderstanding. I presume that the image of the embedding of $N$ will (or at least might) be contained in a ball in $\mathbb{R}P^j$ . Given this, you get a map 2-1 from $M$ to $\mathbb{R}P^j$ which lifts to a map 2-1 to S^j$. Does this make sense?

Comment by Alessandro Sisto

2011-01-11T16:38:38Z

Indeed it is not true in general in that case. I'm sorry, the definition of hyperbolic manifold I implicitly used, and that I typically work with, is "quotient of the hyperbolic space wrt to..." (that is, complete hyperbolic manifold in your sense). In the non-complete case there are indeed many more examples.